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A 0.3146-g sample of a mixture of NaCl(s) and KBr(s) was dissolved in water. The resulting...

A 0.3146-g sample of a mixture of NaCl(s) and KBr(s) was dissolved in water. The resulting solution required 39.10 mL of 0.08765 M AgNO3(aq) to precipitate the Cl–(aq) and Br–(aq) as AgCl(s) and AgBr(s). Calculate the mass percentage of NaCl(s) in the mixture.

PLEASE explain your thought process. I would like to understand how to get to the answer. Thank you!

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Homework Answers

Answer #1

Let the sample contains x g of NaCl.

The mass of KBr will be (0.3146 - x) g

The molar mass of NaCl is 58.44 g/mol and the molar mass of KBr is 119 g/mol

The number of moles of NaCl present are

The number of moles of KBr are

Total number of moles ......(1)

This is also equal to the number of moles of silver nitrate required.

Note: 1 mole of NaCl will react with 1 mole of silver nitrate and 1 mole of KBr will react with 1 mole of silver nitrate.

Hence, total number of NaCl and KBr is also equal to the total number of moles of silver nitrate.

Total number of moles of present in 39.10 mL of 0.08765 M solution is ......(2)

But (1) = (2)

Mass of NaCl present in the sample g

Mass percent of NaCl in the sample %

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